Let me point out that the equivalence mentioned in Simon Henry's answer is not correct. (*Edit.* It's now corrected.)

There is a unique map of sets $\{1\} \to \{1\}$, but there are two cogroup homomorphisms $F(\{1\}) \to F(\{1\})$, namely $x \mapsto x$ and $x \mapsto 1$. Even more trivially, there is *no* map $\{1\} \to \emptyset$, but there is (exactly) one cogroup homomorphism $F(\{1\}) \to F(\emptyset)$.

This is a good lesson that a category is more than just its objects.

The category of cogroups in $\mathbf{Grp}$ is equivalent to $\mathbf{Set}_*$. The equivalence maps a pointed set $(X,x_0)$ to the free group on $X$ modulo $\langle \langle x_0 \rangle \rangle$ (which is isomorphic to the free group on $X \setminus \{x_0\}$, but with this POV the functoriality is less clear) equipped with the usual cogroup structure. This is explained (in a much larger framework) in Bergman's [*An Invitation to General Algebra and Universal Constructions*][1] in Section 10.7. I also mention this in Example 6.1 of my paper on [limit sketches][2].

So it remains to recover $\mathbf{Set}$ from $\mathbf{Set}_*$. This can be done:

$\mathbf{Set}$ is equivalent to the subcategory of $\mathbf{Set}_*$ which has the same objects but only those morphisms of pointed sets $f : X \to Y$ such that

$\begin{array}{ccc} \ast & \xrightarrow{~~} & \ast \\ \downarrow && \downarrow \\  X & \xrightarrow{~f~} & Y  \end{array} $

is a pullback, where $\ast$ is the zero object of $\mathbf{Set}_*$.


  [1]: https://math.berkeley.edu/~gbergman/245/
  [2]: https://arxiv.org/abs/2106.11115